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MATH DAY 2010 at FAU
Competition A–Individual
NOTE:
1. In the multiple choice questions, the option NA stands for “None of the previous
answers is correct.”
2. In all questions, i stands for the imaginary unit; i2 = −1.
3. logb a denotes the logarithm in base b of a; logb a = c if and only if bc = a.
4. If n is a non-negative integer, then n! stands for the product of all positive integers in the range 1 − n if n ≥ 1, with 0! defined to be 1. That is:
0! = 1, 1! = 1, 2! = 2, 3! = 2·3 = 6, 4! = 2·3·4 = 24, 5! = 2·3·4·5 = 120, etc.
5. Do NOT assume that pictures are drawn to scale. They are merely intended as a
guide.
THE ANSWERS
1. The least number of students that must be in a classroom to ensure that there are
at least 10 boys or at least 10 girls is
(A) 10
(B) 11
(C) 19
(D) 20
(E) NA
Solution. With 19 students, if there are less than 10 students of one sex, there
have to be at least 10 of the other. If there are less than 10 girls and less than 10
boys, then there are at most 9 + 9 = 18 students.
The correct solution is C.
2.⋆ If greeting cards cost $2.50 for a box of 12, $1.25 for a packet of three, or 50
cents each, determine the greatest number of cards that can be purchased for
$14.75.
Solution. We can buy 5 boxes, one packet and 2 cards, for a total of 5 × 12 +
3 + 2 = 65 cards.
The correct solution is 65.
3. A store cuts the price of an article by 25%. To restore its price to its original
value, the store must increase the price by:
(A) 24 32 %
(B) 25%
(C) 30%
(D) 33 31 %
(E) NA
Solution. Let us say that the original price is P . After the cut, the price is P −
.25P = .75P . If we now increase this price by a percentage x, the price becomes
x
)0.75P ; equating to P and solving for x we get x = 100[(1/.75) − 1] =
(1 + 100
33.333 . . ..
The correct solution is D.
4. Adam ran up a hill at a speed of 3 miles per hour, and then ran down at a speed
of 6 miles per hour. His average speed for the round trip was (in miles per hour)
(A) 4
(B) 4.5
(C) 5
(D) 5.5
(E) NA
Solution. Let us call L the distance (in miles) to the top of the hill. The time
it took Adam to reach the top was L/3 hours. It then only took him L/6 hours
to get back to the bottom of the hill, so the total time for the round trip was
(L/3) + (L/6). The total distance for the round trip being 2L, his average speed
was
2L
2 × 3 × 6× 6 L
36
=
= 4.
=
L
L
(3
+
6)
6
L
9
+
3
6
(Incidentally, given two positive numbers a, b, the number c = 2ab/(a + b)
which has the property that its reciprocal 1/c is the mean between the reciprocals
1/a, 1/b of a, b, is called the harmonic mean between a and b. The average
velocity in this case is the harmonic mean of the two velocities.)
The correct solution is A.
5. The picture below shows grid lines in the plane, all grid lines intersect at points
with integer coordinates. The lower left corner of the picture is the origin, the
point at (0, 0), the point in the upper right corner has coordinates (20, 10). We
will call a path from the origin to the point (20, 10) proper if it stays on the grid
lines and has a total length of 30. This means that a proper path can only go up
or to the right, never down or to the left. One such path is shown with thick lines.
How many proper paths are there from (0, 0) to (20, 10)?
(20, 10)
b
b
(0, 0)
(A)
30!
10!20!
(B)
30!
10!
(C)
30!
20!
(D)
10!30!
20!
(E) 30!
(F) NA
Solution. The trick here is to realize that every path consists of twenty moves
of one unit to the right and ten moves of one unit up, in some order. If we write
r for a move of one unit to to the right, u for a move of one unit up, every path
can be seen as string of length 30 of r’s and u’s, with exactly 20 r’s and 10 u’s.
For example the path shown in the picture can be coded as
rrrruuurrrrrrrrrrrrruurrruuuuu
There are u’s precisely in positions 5, 6, 7, 21, 22, 26, 27, 28, 29, 30. It is now
easy to see that we get a path precisely for every subset of ten elements of the set
{1, . . . , 30}; we assign to each number in that set the letter u, to each number in
2
the complement of this set the letter r. The number of subsets of 10 elements of
a set of 30 elements is
30!
30
=
.
10
10!20!
The correct solution is A.
6. The number 10! = 3, 628, 800 ends in two zeros. In how many zeros does 500!
end.
Solution. We can factor 500! = 2e 5f a, where e, f are positive integers and
a is an odd number not divisible by 5. It is sort of obvious that e > f so that
there will be exactly f trailing 0’s. (Every 5 can be multiplied by a 2 to produce
a factor of 10.) There are 100 multiples of 5 in the range 1-500. Of these 20 are
also multiples of 25; each one of these adds another factor of 5. Of these 20, 4
are also multiples of 125, adding one more 5. There are thus 100 + 20 + 4 = 124
trailing zeros.
The correct solution is 124.
7. For a certain integer n, the numbers 5n + 16 and 8n + 29 have a common factor
larger than one. That common factor is:
(A) 11
(B) 13
(C) 17
(D) 19
(E) NA
Solution. The common factor has to divide also 5(8n+29)−8(5n+16) = 17.
The factor must be 17.
Another way of coming up with the answer is by trial and error, writing out
values of the two expressions. Beginning with n = 1, 2,, etc., for n = 7, one
gets the first case of a common factor; 5 · 7 + 16 = 51 = 3 · 17 and 8 · 7 + 29 =
85 = 5 · 17.
The correct solution is C.
8. In how many ways can one buy 44-cent and 90-cent stamps with exactly 50
dollars?
(A) 1
(B) 2
(C) 3
(D) 4
(E) NA
Solution. Suppose we buy x 44-cent stamps and y 90-cent stamps. Then
44x + 90y = 5000. Equations of the form ax + by = c have integer solutions
if and only the greatest common divisor g of a, b divides c and then all solutions
are of the form x = x0 + (b/g)k, y = y0 − (a/g)k, where (x0 , y0 ) is a particular
solution and k = 0, ±1, ±2, . . .. In our case a = 44, b = 90, so g = 2, which
divides 5000. There are several ways of finding a particular solution (using the
Euclidean algorithm, for example), but guessing and/or trial and error can be
as good as any. For example, we might notice that two times 44 is 88, that
is 44(−2) + 90(1) = 2, from which 44(−5000) + 90(2500) = 5000. Thus
one solution is (x0 , y0 ) = (−5000, 2500). Of course, this solution would mean
buying a large negative number of stamps, so it does not apply. But we now
know that all solutions are of the form x = −5000 + 45k, y = 2500 − 22k, k
an integer. To get x to be non-negative we need to have k ≥ 5000/45; that is
k ≥ 112; to get y ≥ 0 we need k ≤ 2500/22; that is, k ≤ 113. For k = 112
we get (x, y) = 40, 36); for k = 113 we get (x, y) = (85, 14). In other words
there are two ways of using up the 50 dollars. We can buy either forty 44-cent
3
stamps and thirty six 90-cent stamps, or we can buy eighty five 44-cent stamps
and fourteen 90-cent stamps.
The correct solution is B.
9. Determine the two last digits of 32010 .
(A) 03
(B) 21
(C) 27
(D) 49
(E) 81
(F) NA
Solution. We work modulo 100. One approach is to try to determine a pattern
among the last two digits as we keep multiplying by three. If we do this, retaining
only the last two digits we get the following sequence:
03, 09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01
and from now on everything repeats. We have a cycle of length 20. The closest
multiple of 20 to 2010 is 2000, so that now, working mod 100 and using ≡ to
mean equivalence mod 100 (same last two digits):
32010 = 32000 310 ≡ 1 × 49 = 49
(49 being the tenth entry in our list).
One can shorten this process significantly using Euler’s extension of Fermat’s
little theorem; namely that if n is a positive integer and a is an integer relatively
prime with n, then aφ(n) ≡ 1 (mod n), where
1
1
φ(n) = n 1 −
··· 1 −
p1
pr
is the so called Euler φ function of n, and p1 , . . . , pr are all the distinct prime
factors of n. Then φ(100) = 100(1 − 12 )(1 − 51 ) = 40 and we would have started
with 340 ≡ 1 (mod 100).
The correct solution is D.
10. If loga x = 28 and logb x = 14, then logb a equals
(A)
1
2
(B) 2
(C) 42
(D) 24
(E) Can’t be determined
(F) NA
Solution. loga x = 28, logb x = 14 means that x = a28 = b14 ; solving
a = b14/28 = b1/2 . Taking logarithm in base b, logb a = 1/2.
The correct solution is A.
11. Let f (x) be a function such that f (x) + 2f (−x) = sin x for every real number
x. What is the value of f π2 ?
(A) − 1
(B) −
1
2
(C)
1
2
(D) 1
(E) NA
Solution. Replacing x by −x in the equation for f we get f (−x) + 2f (x) =
− sin x. Together with the original equation f (x) + 2f (−x) = sin x we get a
system of two equations and we can solve for f (x) to get f (x) = − sin x. Thus
f (π/2) = − sin(π/2) = −1.
The correct solution is A.
4
4
12. Determine the coefficient of x5 if the expression (1 + x2 )3 − 2x is expanded
and written in standard polynomial form.
(A) 0
(B) − 187
(C) − 300
(D) − 384
(E) NA
Solution.
4
(1 + x2 )3 − 2x = (1+x2 )12 −8x(1+x2 )9 +24x2 (1+x2 )6 −32x3 (1+x2 )3 +16x4 .
Of these terms, only −8x(1 + x2 )9 and −32x3 (1 + x2 )3 will contribute a fifth
power of x. If we remember our binomial expansions one sees that the total
contribution is
9
3
−8
− 32
= −384.
2
1
The correct solution is D.
13. Suppose it is known that one root of the equation 3x2 + ax + b = 0, where a, b
are real numbers, is 2 + 3i. The value of b is
(B) − 3
(A) Undetermined
(C) 13
(D) 26
(E) 39
(F) NA
Solution. Because all coefficients are real, if a complex number solves the
equation, so does its conjugate. It follows that the roots of the equation are
2+3i, 2−3i. Since the leading coefficient is 1, we have b/3 = (2+3i)(2−3i) =
13.
The correct solution is E.
14. The remainder of dividing x33 − 4x31 + 32x − 7 by x − 2 is
(A) 0
(B) 231
(C) − 7
(D) 57
(E) 64
(F) NA
Solution. The remainder of dividing a polynomial by a monomial of the form
x − a equals the value of the polynomial at a. In our case the remainder is
233 − 4 · 231 + 32 · 2 − 7 = 57.
The correct solution is D.
15. Tom has 13 numbers. He adds up all possible products of the numbers and finds
a sum S. Now, he adds a 14-th number to the list, adds up all possible products
of the numbers, and finds a sum T . What is the 14-th number?
(A)
T
S
(B)
1+T
1+S
(C)
T −S
1+S
(D)
T +S
S−1
(E) NA
Note. Among “all possible products” we have to include products of a single
factor; that is, the 13 numbers themselves so that if the numbers are a1 , . . . a13 ,
then
S = a1 + · · · + a13 + a1 a2 + · · · + a12 a13 + a1 a2 a3 + · · · + a1 a2 · · · a13 .
Solution. Suppose the fourteenth number is x. Then the possible products
consist of all the previous possible products (which add up to S), plus x, plus
every previous product multiplied by x (which adds up to Sx). Thus T = S +
x + Sx so that x = (T − S)/(1 + S)
The correct solution is C.
5
16. Determine the largest number n such that 1 ≤ n ≤ 1000 having an odd number
of positive divisors.
(A) 789
(B) 841
(C) 900
(D) 999
(E) NA
Solution. Divisors come in pairs; if a divides n so does n/a. The only case
in which one has an odd number of divisors is when there is a divisor a such
that a = n/a; that is, when n is a perfect square. We thus have to determine the
largest perfect square ≤ 1000, which is easily calculated as being 961.
The correct solution is E.
17.⋆ How many positive integers divide at least one of the following two numbers:
310 · 3520 , 615 · 510 ?
(Just to make sure, the first number is three to the power ten times thirty five to
the power twenty; the second number is six to the fifteen times five to the ten.)
Solution. The number we are looking for is the total number of divisors of
310 3520 , plus the total number of divisors of 615 510 , minus the number of common divisors.
The total number of positive divisors of a number whose prime factorization
is pe11 · · · penn , where p1 , . . . , pn are distinct primes and e1 , . . . , en are positive
integral powers, is (e1 + 1) · · · (en + 1). Thus 310 3520 = 310 520 720 has a total
of 11 × 21 × 21 = 4851 divisors, 615 510 = 215 315 510 has 16 × 16 × 11 =
2816 divisors. A positive divisor of both numbers must have the form 3e 5f ,
where 0 ≤ e ≤ 10, 0 ≤ f ≤ 11, a total of 11 × 11 = 121 numbers. Now
4851 + 2816 − 121 = 7546.
The correct solution is 7491.
18.⋆ Determine the smallest positive number with exactly 56 divisors.
Solution.
If the prime factorization of a number n is of the form
n = pe11 pe22 · · · perr
where p1 , . . . , pr are distinct primes and e1 , . . . , er are positive integers, then the
divisors of n are all (and no more than) the numbers of the form
pf11 pf22 · · · pfrr
where 0 ≤ fk ≤ ek for k = 1, . . . , r; a total of (e1 + 1)(e2 + 1) · · · (er + 1)
numbers. If a number is to have 56 divisors, then we must have (e1 + 1)(e2 +
1) · · · (er + 1) = 56. The possible factorizations of 56 are (not ignoring the case
r = 1).
56 =
56,
56 =
7 · 8,
=
=
2 · 4 · 7,
2 · 2 · 2 · 7,
We try to get the smallest number corresponding to each one of these. Concerning the first factorization (r = 1), we have a single prime involved with
a power of e1 = 55; the smallest number with 56 divisors and a single prime
factor is 255 , a pretty large number. Chances are we won’t need to compute its
6
exact value. The factorization 56 = 7 · 8 involves two prime factors; we’d get
n = p61 p72 ; obviously the smallest such number is 36 27 = 93, 312. The factorization 56 = 2 · 4 · 7 corresponds to numbers of the form p1 p32 p63 ; the smallest
such number is 5 · 33 · 26 = 8640. Finally, the factorization 56 = 2 · 2 · 2 · 7 has
as smallest number the number 7 · 5 · 3 · 26 = 6720.
The correct solution is 6720.
19. The equation
x5 + ax4 + bx3 + cx2 + dx + e = 0
has the roots 1, 2, 3, 4, and 5. Determine c.
(A) 0
Solution.
(B) − 121
(C) − 200
(D) − 220
(E) − 225
(F) NA
By Viète’s relations,
c = −(1·2·3+1·2·4+1·2·5+1·3·4+1·3·5+1·4·5+2·3·4+2·3·5+2·4·5+3·4·5) = −225
The correct solution is E.
20. Suppose it is known that an equation of the form
xn + an−1 xn−1 + · · · a1 x + 10 = 0,
where all coefficients an−1 , an−2 , . . . , a1 are integers, has four distinct positive
integer roots m1 , m2 , m3 and m4 . Determine m1 + m2 + m3 + m4 .
(A) 10
Solution.
(B) 18
(C) 20
(D) 30
(E) 36
(F) NA
If m > 0 is an integer root of the equation then
mn +an−1 mn−1 +· · · a1 m+10 = 0, thus m mn−1 + an−1 mn−2 + · · · a1 = −10
and it follows that m divides 10. Now 10 has exactly four positive divisors,
namely 1, 2, 5 and 10 so if the equation has four distinct positive integral roots,
they must be 1, 2, 5, 10, which add up to 18.
The correct solution is B.
21. Let x, y, z be real numbers such that x + y + z = 1 and
1
1 1
+ + = 1. Then
x y z
(A) They are all different, and none of them is 1. (B) All three of them must coincide.
(C) One of them must be 1. (D) Two of them must equal 1. (E) NA
Solution. Another way of writing the equations satisfied by x, y, z is x + y +
z = 1, xy + yz + zx = xyz, which suggests that x, y, z are roots of an equation
X 3 − X 2 + aX − a = 0. This equation clearly has a root equal to 1. It suggests
doing the product (1 − x)(1 − y)(1 − z). One gets
(1 − x)(1 − y)(1 − z) = 1 − (x + y + z) + (xy + xz + yz) − xyz = 0.
This proves that at least one of x, y, z must equal 1; say x = 1. Then y = −z
and 1/y = −1/z (which is the same as y = −z), in particular y 6= z (neither y
nor z can be 0). At this point we know that the correct answer is either C or D.
The key word now is must. We can have two of the numbers equal 1, but that is
not a necessity. For example, x = 1, y = 2, z = −2 is a possible solution.
The correct solution is C.
7
22. Two congruent rectangles of dimensions a×b share a common diagonal as shown
in the diagram below. Find the area of their overlap.
b
a
(A)
a(a2 + b2 )
b
Solution.
(B)
b(a2 + b2 )
a
(C)
a(a2 + b2 )
2b
(D)
b(a2 + b2 )
2a
(E) NA
Let us add a few labels and lines to the figure for easy reference.
E
C
B
D
b
A
F
a
Now a = |BD|, b = |AB|. It is easy to see that △ABC is congruent to △DEC
and the area of the shaded region is |CD| × |CF | = |CD| × b. Noticing that
|CE| = a − |CD| we get by Pythagoras
|CD|2 = b2 +|CE|2 = b2 +(a−|CD|)2 ; solving for |CD|,
The area is thus |CD| × b =
|CD| =
a2 + b 2
.
2a
b(a2 + b2 )
.
2a
The correct solution is D.
23.⋆ Two poles, one 30 feet tall, the other one 45 feet tall, are set up near each other.
From the top of each pole a wire is stretched to the base of the other pole. How
high above the ground do the two wires meet (intersect)? (Determine x in the
picture below.)
45 ft
30 ft
x
8
Solution. Let the poles be at points A, B and let C be the point on the ground
above which the two wires meet, as in the picture below. Let M be the point
where the wires meet, so x = |CM |. Let also D, E be as in the picture.
E
D
M
30 ft
45 ft
x
A
C
B
By similarity of triangles △CBM ∼ ABD,
△CAM ∼ △BAE,
x
|AC|
=
. Adding:
45
|AB|
x
|CB|
=
. By the similarity
30
|AB|
x
x
|CB| |AC|
|CB| + |AC|
+
=
+
=
= 1.
30 45
|AB| |AB|
|AB|
Solving now for x, x = (30 · 45)/(30 + 45) = 18.
The correct solution is 18.
24. A circle is drawn through vertices A and D of a square ABCD in such a way
that the circle is tangent to side BC.
D
C
A
B
If the length of a side of the square is ℓ, the radius of the circle equals
ℓ
(A)
2
√
ℓ 2
(B)
2
5ℓ
(C)
8
√
5ℓ 2
(D)
8
(E)
3ℓ
8
(F) NA
Solution. Consider the shaded triangle in the picture below. It is a right triangle; one leg, the horizontal one, goes from the center of the circle to the side of
the square and has length ℓ − r, the vertical leg has length ℓ/2. The hypotenuse
coincides with the radius of the circle.
9
D
C
A
B
By Pythagoras: r2 = (r − ℓ)2 +
ℓ2
. Solving for r, r = 5ℓ/8.
4
The correct solution is C.
25. Between two congruent circles that are tangent to each other at A and to a line ℓ
at B, C, a third circle is constructed tangent to both circles and also to ℓ. What
is the ratio of the area of the small circle to the area of the curvilinear “triangle”
ABC?
A
ℓ
B
(A)
1
2
(B)
C
π
8(4 − π)
(C)
π
8(4 + π)
(D)
1
3
(E) NA
Solution. The area of the curvilinear triangle ABC is equal to the area of
the rectangle of vertices B, C and the centers of the two congruent (big) circles
πr2
. It is
minus the area of two quarters of these circles; that is, it equals 2r2 −
2
easy to see that the radius ρ of the third (small) circle is r/4. For example, if we
think of A as being the origin, then the center of the small circle has coordinates
(0, −(r − ρ), center of the large circle on the right has coordinates (r, 0). The
square of the distance between these two centers is thus r2 + (r − ρ)2 ). On the
other hand, because these circles are tangent, the distance is the sum of the radii.
Thus r2 + (r − ρ)2 ) = (r + ρ)2 from which ρ = r/4 follows. The area of the
small circle is thus πr2 /16 and the answer to the question is
πr 2
16
2r2
−
πr 2
2
The correct solution is B.
10
=
π
.
8(4 − π)